The fact that Ramanujan’s Constant eπ√163 is almost an integer (262537412640768743.99999999999925…) doesn’t seem to be a coincidence, but has to do with the 163 appearing in it. Can you explain why it’s almost-but-not-quite an integer in layman’s terms (I’m not a mathematician)?
Answer
This is quite a challenge to express in “layman’s terms”, but the
reason is that
j(1+√−1632)
is an integer where j is the j-function. When you substitute
(1+√−163)/2 into the q-expansion (see the wikipedia page)
of j, all terms save the first two are small, and the first two equal
−exp(π√163)+744.
The reason that this j-value is an integer is due to the quadratic
field Q(√−163) having
class number one, or equivalently that all positive-definite
integer binary quadratic forms of discriminant −163 are equivalent.
Added
I’ll try to explain the connection with binary quadratic forms. Consider
a quadratic form
Q(x,y)=ax2+bxy+cy2
with a, b and c integers. I’ll only consider forms Q which are
primitive, so that a, b and c have no common factor >1,
and positive-definite, that is a>0 and the discriminant
D=b2−4ac<0. There is a notion of equivalence of quadratic forms,
and two primitive positive-definite forms Q and Q′(x,y)=a′x2+b′xy+c′y2
(necessarily also of discriminant D) are equivalent if and only if
j(b+√−D2a)=j(b′+√−D2a′).
For each possible discriminant there are only finitely many equivalence
classes. Thus we get a finite set of j-values for each discriminant, and
the big theorem is that they are the solutions of a monic algebraic equation
with integer coefficients. When there is only one class the equation has the
form x−k=0 where x is an integer, and the j-value must be an integer.
My recommended reference for this is David Cox's book
Primes of the form x2+ny2. But these results appear towards
the end of this 350-page book.
Attribution
Source : Link , Question Author : stevenvh , Answer Author : Robin Chapman